cm 197 mechanics of materials chap 13 shear forces and bending moments in beams l.
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CM 197 Mechanics of Materials Chap 13: Shear Forces and Bending Moments in Beams. Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials , 2 nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997) . CM 197. Chap 13: Shear Forces and Bending Moments. Topics

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CM 197 Mechanics of Materials Chap 13: Shear Forces and Bending Moments in Beams


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cm 197 mechanics of materials chap 13 shear forces and bending moments in beams
CM 197Mechanics of Materials Chap 13: Shear Forces and Bending Moments in Beams

Professor Joe Greene

CSU, CHICO

Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997)

CM 197

chap 13 shear forces and bending moments
Chap 13: Shear Forces and Bending Moments
  • Topics
    • Introduction
    • Types of Beams
    • Types of Loading
    • Beam Reactions
    • Shear Force and Bending Moment in Beams
    • Shear Force and Bending Moment Diagrams
    • Relationships Among Load, Shear, and Moment
    • Sketching Shear and Moment Diagrams Using Relationships
    • Shear and Moment Formulas
introduction
Introduction

Plane of symmetry

  • Beams are members that carry transverse loads and are subjected to bending.
    • Any structural member subjected to bending may be referred to as a beam.
      • Shafts, girders, stringers, floor beams, joists, etc.
  • Objective of chapter
    • Determine the internal forces at various sections along a beam.
      • Review Types of beams and supports.
      • Calculation of beam sections.
      • Calculation of internal forces (shear and moments) in the beam.
      • Produce graphical representations of shear and bending moments.
  • Assumptions
    • Beams are straight and of uniform cross sections that possess a vertical plane of symmetry. Fig 13-1
    • Horizontal positions
    • Subjected to forces applied in the vertical plane of symmetry.
types of supports for beams
Types of Supports for Beams

R

Rx

Ry

Rx

M

Ry

  • Supports
    • Roller supports
      • Roller or link support resists motion of the beam only along the direction perpendicular to the plane of the support (along axis)
        • Reaction at roller support acts along the known direction. Fig 13-2
        • Have one unknown Reaction force, R
    • Hinge supports
      • Resists motion of the beam at the support in any direction on the plane of loading.
      • Have two unknown Reaction forces- Rx and Ry
    • Fixed supports
      • Beam is built into wall or column.
      • End of beam is fixed and doesn’t move.
      • Have two unknown Reaction forces, Rx and Ry, and one unknown Moment, M
types of beams
Types of Beams
  • Beams
    • Simple Beam (Statically Determinate- 3 unknowns)
      • Beam supported at its end with a hinge and a roller
    • Overhanging Beam (Statically Determinate- 3 unknowns)
      • Simple supported beam with an overhang from one or both ends
    • Cantilever Beam (Statically Determinate- 3 unknowns)
      • Beam that is fixed at one end and free at the other
    • Propped cantilever beam (Statically Indeterminate >3 unknowns)
      • Beam that is fixed at one end and simply supported at the other.
    • Fixed Beam (Statically Indeterminate >3 unknowns)
      • Beam with both ends are fixed to supports
    • Continuous Beam (Statically Indeterminate >3 unknowns)
      • Beam is supported on a hinge and two or more roller supports
types of loading
Types of Loading
  • Loading
    • Concentrated Loads
      • Load is applied at a specific point on the beam and is considered a discrete force acting at the point.
        • Example, weight fastened to a beam by a cable that applies a concentrated load.
    • Uniform Loads
      • Load is distributes over a part or the entire length of the beam.
        • Load is force per unit length of beam (lb/ft or N/m)
        • Replaced with equivalent force = uniform load times length of beam
          • Example, weight of beam.
    • Linearly Varying Loads
      • Load is distributed with a uniform variation of intensity.
        • Occurs on a vertical or inclined wall due to liquid pressure.
        • Load changes along the length of the beam.
beam reactions
Beam Reactions
  • Beam Reactions
    • Review of Chapter 3
    • Procedure
      • Write three independent equilibrium equations to solve for the three unknowns.
      • Three equations could be combinations of force-component or moment equations, as long as they are independent of each other.
      • This chapter assume no horizontal forces.
        • Thus reaction force in horizontal direction is zero.
      • Assume, sum of forces in x direction is 0, thus the horizontal reaction is 0.
    • Example, 13-1
    • Example, 13-2
shear force bending moment in beams
Shear Force & Bending Moment in Beams

100 lb

200 lb

Rx

1 ft

3 ft

Ry

R

10 ft

  • Shear Force and Bending Moment
    • Developed in a beam to resist the external forces and to maintain equilibrium.
    • Figure 13-7
      • Beam is 10 ft long with roller support on the right and hinge support on left.
      • Beam is subjected to two concentrated loads.
      • Results
        • For cross section at 3 feet from the left, the forces and moments need to be balanced.
        • Balance the forces and moments at each section
          • Normal forces: Rx = 0; Ry =150 pounds;
          • Shear force: V = 150 – 100 = 50 lbs; Moment is acting clockwise and = 3*150 – 100*2 = 250lb-ft;
          • Internal resisting moment moment has to be equal and opposite of moment and acting opposite direction. M = 250ft-lb counterclockwise.
        • For results for cross section from the right, the forces and moments are calculated in the same manner.
beam sign conventions
Beam Sign Conventions
  • Positive Shear
      • Shear force at a section is positive if the external forces on the beam produce a shear effect that ends to cause the left side of the section to move up relative to the right side. Fig 13-8a
  • Positive Moment
      • Bending moment at a section is positive if external forces on the beam produce a bending effect that causes the beam to concave upward at the section. Fig 13-8b
    • Positive internal shear force, V, at a given section of a beam viewed from both directions is shown in Fig 13-8c.
    • Positive internal moment M at a given section of a beam viewed from both directions is shown in Fig 13-8d.
  • Rule 1 for finding shear forces
      • Internal shear force at any section of a beam is equal to the algebraic sum of the external forces on either segment separated by the section.
        • From the left of beam: upward force as positive
        • From the right of the beam: downward force as positive
  • Rule 2 for finding bending moments
      • Internal bending moment at any section of a beam is equal to the algebraic sum of the moments about a section due to external forces.
        • From either side: moment due to upward force as positive
    • Example, 13-1 and Example, 13-2
shear force and moment diagrams
Shear Force and Moment Diagrams
  • Shear force and bending moment diagrams depict the variation of shear force and bending moment along a beam.
    • Help visualize the shear forces and bending moments along beam
    • Construct diagram
      • Ordinate (y-axis) values are from balance of forces and moments.
      • Compute reaction forces and moment couples.
      • Plot values of the shear just below the free body diagram on the y-axis with positive values above the baseline and negative values below the baseline.
      • Plot values of moments just below the shear diagram on same x-axis scale with positive values above the baseline and negative values below the baseline.
  • Important aspects of diagrams
    • For concentrated loads
      • The shear diagram has a horizontal line between the loads
      • The moment diagram has a line with constant slope
    • For distributed loads
      • The shear diagram has a line with a constant slope.
      • The moment diagram has a curved line between moment points.
    • If the shear diagram crosses the horizontal line (at zero shear) then it has a maximum or minimum moment at that point.
    • Example 13-5
relationships among load shear and moment
Relationships among Load, Shear, and Moment
  • Certain relationships exist among the loading diagram, shear diagram and moment diagram.
  • Assume incremental element of a beam.
    • Relationships between Load and Shear
      • For a distributed load, w, applied to a beam,
        • The slope of the shear diagram (rate of change of the shear force per unit length) at any section is equal to the load intensity at that section.
        • The shear force at a section is equal to the shear force at the previous section plus the total load between the two sections.
      • Shear force at the section immediately to the right of a concentrated load is equal to the shear force immediately to the left plus the load.
        • Shear force has an abrupt change at the concentrated load, upward = +
    • Relationship between shear and moment
      • Moment at a section is equal to the moment at the previous section plus the area under the shear diagram between the two sections.
using relationships of diagrams
Using Relationships of Diagrams
  • Loading Diagram or free body diagram
    • Show all applied forces and reactions on beam plus dimensions
    • Never replace a distributed load with an equivalent concentrated load
  • Shear diagram procedure
    • Draw shear diagram directly below the loading diagram.
      • Horizontal line is drawn at proper location below loading diagram.
      • Draw vertical lines from controlling sections for supports and loads.
    • Start at left end and compute shear at controlling sections using Eqn 13-5 VB = VA + LoadB-A
      • Note: Section with concentrated force is applied, shear force diagram has an abrupt change at concentrated load location.
    • Plot points on shear diagram using forces at each load.
      • Connect points
      • Note: Sections with distributed loads have a sloping line between them.
      • Note: Slope of line is the intensity of the load. Concentrated loads have slope = 0
using relationships of diagrams13
Using Relationships of Diagrams
  • Moment diagram procedure
    • Draw moment diagram directly below the shear diagram.
      • Horizontal line is drawn at proper location below loading diagram.
      • Draw vertical lines from controlling sections for supports and loads
    • Calculate areas from shear diagram for each section.
      • Note: moments at free end or ends of a simple beam are always = 0.
    • Start at the left end and compute the moments using Eqn 13-7, MB = MA + AreaB-A
    • Plot points on moment diagram using previous moments and areas at each section
      • Connect points with straight lines (for concentrated loads) or curves (for distributed loads)
      • Note: Maximum or minimum of moment occurs at the location where the shear is = 0 or where the shear changes sign.
    • Example, 13-6, 13-7, 13-8, 13-9
shear and moment formulas
Shear and Moment Formulas
  • From previous section, shear force and bending moment diagram can be plotted for some simple but typical loading conditions.
    • Table 13-1
  • Procedure
    • Match the loading conditions to one or more loading examples in Table 13-1
      • Get the loading solutions from the example and use it in problem for
        • Reaction forces, Maximum shear, maximum moments, etc.
    • Method of superposition
      • Effect of each load is computed separately and the then combined effect is added algebraically.
    • Example, 13-10, 13-11